Question 1

How many small cubes of edge 1 cm (each of volume 1 cm³) are needed to exactly fill a vessel of volume 1 dm³?

Accepted Answer

1000. Since 1 dm = 10 cm, a volume of 1 dm³ = (10 cm)³ = 1000 cm³ (which also equals 1 L). As each small cube is 1 cm³, exactly 1000 of them fill 1 dm³.

Question 2

Assertion (A): Chemists have successfully synthesised safer alternatives to CFCs (chlorofluorocarbons). Reason (R): CFCs are responsible for ozone depletion in the stratosphere. Choose the correct option.

Accepted Answer

Both A and R are true and R is the correct explanation of A. CFCs deplete stratospheric ozone (R is true), which is precisely why chemists developed safer refrigerant alternatives (A is true). The ozone-depleting nature of CFCs is the reason such alternatives were needed, so R correctly explains A.

Question 3

Which of the following ancient Indian texts is correctly matched with the contribution attributed to it?

Accepted Answer

Kautilya's Arthashastra — production of salt from sea. Kautilya's Arthashastra describes the production of salt from sea — the correct match. Mercury compounds come from Nagarjuna's Rasratnakar (Rasopanishada deals with gunpowder), alkalies from the Sushruta Samhita (the Vaiseshika Sutras give the atomic theory), and the Rigveda deals with tanning and dyeing, not gunpowder.

Question 4

Methane (CH₄) and carbon dioxide (CO₂) are two greenhouse gases. For equal masses of the two gases (C = 12, H = 1, O = 16), the ratio of the number of molecules of methane to carbon dioxide is:

Accepted Answer

11 : 4. Molar masses: CH₄ = 16, CO₂ = 44 g mol⁻¹. For equal mass m, moles of CH₄ = m/16 and moles of CO₂ = m/44. Number of molecules is proportional to moles, so the ratio = (1/16) : (1/44) = 44 : 16 = 11 : 4. Equal masses do NOT give equal molecules (that wrong idea gives 1 : 1); inverting the molar-mass ratio wrongly gives 4 : 11.

Question 5

Consider the following six substances: copper, water, air, oxygen, carbon dioxide and a sugar solution. How many of them are compounds?

Accepted Answer

2. A compound contains two or more different elements in a fixed ratio: only water (H₂O) and carbon dioxide (CO₂) qualify — that is 2. Copper and oxygen are elements (one type of atom each), while air and the sugar solution are mixtures. So exactly two of the six are compounds.

Question 6

The Celsius and Fahrenheit temperature scales are related by °F = (9/5)°C + 32. At what temperature will a thermometer show the same numerical reading on both the Celsius and Fahrenheit scales?

Accepted Answer

−40°. Set the two readings equal: x = (9/5)x + 32 ⇒ x − (9/5)x = 32 ⇒ −(4/5)x = 32 ⇒ x = −40. So −40 °C equals −40 °F — the only temperature that is numerically the same on both scales.

Question 7

Equal masses of two liquids of densities 2 g cm⁻³ and 3 g cm⁻³ are mixed. If the volumes are additive, the density of the mixture is:

Accepted Answer

2.4 g cm⁻³. Take each mass as m. Volumes are m/2 and m/3, so total volume = m/2 + m/3 = 5m/6 and total mass = 2m. Density = 2m ÷ (5m/6) = 12/5 = 2.4 g cm⁻³. It is NOT the simple average 2.5 g cm⁻³, because the two volumes are unequal.

Question 8

The density of a gas is 1.25 g L⁻¹. Using 1 g = 10⁻³ kg and 1 L = 10⁻³ m³, its density in kg m⁻³ is:

Accepted Answer

1.25 kg m⁻³. Convert both units: 1 g L⁻¹ = (10⁻³ kg) ÷ (10⁻³ m³) = 1 kg m⁻³, since the two factors of 10⁻³ cancel. Therefore 1.25 g L⁻¹ = 1.25 kg m⁻³. (The g cm⁻³ → kg m⁻³ conversion multiplies by 1000, but for g L⁻¹ the factor is 1 — a common trap giving 1250.)

Question 9

A metal M forms three oxides containing 40.0%, 50.0% and 57.1% oxygen by mass. For a fixed mass of the metal, the ratio of the masses of oxygen in the three oxides is:

Accepted Answer

2 : 3 : 4. Oxygen per unit mass of metal = %O ÷ %M: for the oxides these are 40/60 = 0.667, 50/50 = 1.000 and 57.1/42.9 = 1.333. Dividing by the smallest (0.667) gives 1 : 1.5 : 2 = 2 : 3 : 4, a simple whole-number ratio (law of multiple proportions).

Question 10

2 g of hydrogen combines completely with 16 g of oxygen to form water. If instead 2 g of hydrogen is mixed with 32 g of oxygen in a closed vessel and ignited, the mass of water formed and the mass of oxygen left unreacted are respectively:

Accepted Answer

18 g of water and 16 g of oxygen. By the law of definite proportions hydrogen and oxygen combine only in the ratio 2 : 16 (i.e. 1 : 8). So 2 g of hydrogen uses just 16 g of oxygen, forming 2 + 16 = 18 g of water (conservation of mass). The extra oxygen, 32 − 16 = 16 g, stays unreacted.

Question 11

At the same temperature and pressure, equal volumes of oxygen (O₂, molar mass 32 g mol⁻¹) and an unknown gas G weigh 1.6 g and 2.2 g respectively. Using Avogadro’s law, the molar mass of G is:

Accepted Answer

44 g mol⁻¹. By Avogadro’s law equal volumes at the same T and P contain equal numbers of molecules (equal moles). Moles of O₂ = 1.6 ÷ 32 = 0.05 mol, so G is also 0.05 mol. Molar mass of G = 2.2 ÷ 0.05 = 44 g mol⁻¹.

Question 12

At constant temperature and pressure, 1 volume of a gaseous compound of nitrogen and hydrogen decomposes to give 0.5 volume of nitrogen and 1.5 volumes of hydrogen. Applying Gay-Lussac’s law and Avogadro’s law, the molecular formula of the compound is:

Accepted Answer

NH₃. Volumes are proportional to numbers of molecules. Doubling to remove the fraction: 2 volumes of compound → 1 volume N₂ + 3 volumes H₂. So 2 molecules of compound contain 2 nitrogen atoms and 6 hydrogen atoms, giving 1 N and 3 H per molecule — the formula NH₃.

Question 13

Assertion (A): Dalton’s atomic theory could not explain Gay Lussac’s law of combining gaseous volumes. Reason (R): Dalton believed that atoms of the same element could not combine, so molecules such as H₂ and O₂ did not exist.

Accepted Answer

Both A and R are true, and R is the correct explanation of A. Both are true. Because Dalton rejected molecules made of like atoms (R), he could not accept that two volumes of hydrogen combine with one volume of oxygen to give two volumes of water vapour; only Avogadro’s polyatomic molecules explained this. Hence R is the correct reason that A holds.

Question 14

Assertion (A): The discovery of isotopes did not force scientists to abandon Dalton’s atomic theory entirely. Reason (R): Only the postulate that all atoms of an element are identical in mass had to be modified, while the central idea that atoms are the units of chemical combination was retained.

Accepted Answer

Both A and R are true, and R is the correct explanation of A. Both are true. Isotopes clash only with the "identical mass" postulate; the rest of Dalton’s framework — atoms as the building blocks that combine in fixed ratios — survives. Because the theory needed modification rather than rejection (R), it was not discarded (A), so R correctly explains A.

Question 15

Despite its later-recognised limitations, Dalton’s atomic theory is regarded as a landmark in chemistry mainly because it:

Accepted Answer

provided a single theoretical model that accounted for the known laws of chemical combination. Dalton’s achievement was to unify the empirical laws of chemical combination (conservation of mass, definite and multiple proportions) under one atomic model. It said nothing correct about atomic structure, bonding, isotopes or isobars — those understandings came much later.

Question 16

A sample of carbon dioxide obtained from burning coal and another obtained from fermentation are both found to contain carbon and oxygen in the same 3:8 mass ratio. Which postulate of Dalton’s theory most directly accounts for this?

Accepted Answer

Compounds are formed when atoms of different elements combine in a fixed ratio. If CO₂ is always built from carbon and oxygen atoms in the same fixed ratio (1 C : 2 O, i.e. 12:32 = 3:8 by mass), then every pure sample must show that same composition regardless of source. This "fixed ratio" postulate is the atomic-theory basis of the law of definite proportions.

Question 17

An element X forms two oxides. The first contains 20.0% oxygen by mass and the second contains 33.3% oxygen by mass. For a fixed mass of X, the ratio of the mass of oxygen in the second oxide to that in the first is:

Accepted Answer

2:1. For a fixed mass of X, oxygen combined per unit X = (%O)/(%X). First oxide: 20/80 = 0.25; second oxide: 33.3/66.7 = 0.50. Ratio (second:first) = 0.50:0.25 = 2:1, a simple whole-number ratio consistent with the law of multiple proportions. Comparing the oxygen percentages directly (33.3:20 ≈ 5:3) is wrong because it ignores that the mass of X must be held constant.

Question 18

Lithium exists as two isotopes, ⁶Li and ⁷Li, and its average atomic mass is 6.94 u. The percentage abundance of the ⁷Li isotope is:

Accepted Answer

94%. Let the fraction of ⁶Li be x. Then 6x + 7(1 − x) = 6.94 ⇒ 7 − x = 6.94 ⇒ x = 0.06. So ⁶Li is 6% and ⁷Li is 1 − 0.06 = 0.94, i.e. 94%. The 6% option is the abundance of the lighter isotope, not the ⁷Li asked for.

Question 19

Magnesium occurs as three isotopes: ²⁴Mg (79%), ²⁵Mg (10%) and ²⁶Mg (11%), with isotopic masses 24, 25 and 26 u. Its average atomic mass is:

Accepted Answer

24.32 u. Average atomic mass = (0.79 × 24) + (0.10 × 25) + (0.11 × 26) = 18.96 + 2.50 + 2.86 = 24.32 u. A plain average of 24, 25 and 26 gives 25.00 u, and using only the most abundant isotope gives 24.00 u; both ignore the actual abundances.

Question 20

A metal hydroxide X(OH)₂ has a formula mass of 74 u (O = 16 u, H = 1 u). The atomic mass of the metal X is:

Accepted Answer

40 u. X(OH)₂ contains one X, two O and two H atoms: mass = X + 2(16 + 1) = X + 34. So X = 74 − 34 = 40 u (the metal is calcium, giving Ca(OH)₂). Forgetting the two hydrogens (X + 32 = 74) gives 42 u; using only one hydroxide group (X + 17 = 74) gives 57 u; 34 u is just the mass of the two OH groups.

Some Basic Concepts of Chemistry — IISER Chemistry MCQs with Solutions

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Sample questions with solutions